wouldn't it be easier to just say that x,y are in the domain of the operators A, B (which i agree IS important to state, as is actually applying the operator to an element as opposed to treating it as some "algebraic" quantity as many books do)? in which case the proof would reduce to about 4 lines. in other words, when would the domain of A and its hermitian conjugate NOT be the same?

The spectrum of a hermitean operator is not necessarily real. Actually only the eigenvalues (points in the pure point spectrum) are real, however, one cannot guarantee that [itex] |n\rangle [/itex] is an element of the Hilbert space. Therefore one cannot guarantee that "a_{n}b_{n}" is real. And even if it was real, there are examples of nonhermitean operators with real spectral values.

The spectrum of a hermitean operator is not necessarily real. Actually only the eigenvalues (points in the pure point spectrum) are real, however, one cannot guarantee that [itex] |n\rangle [/itex] is an element of the Hilbert space. Therefore one cannot guarantee that "a_{n}b_{n}" is real. And even if it was real, there are examples of nonhermitean operators with real spectral values.

Dear dextercioby, At alisa's level I thought it might go like this:

Let

(1) A|x> = a|x>. If A is hermitean then (by definition):

(2) <x|A|x> = <x|A|x>*. Substituting (1) into (2):

(3) <x|a|x> = <x|a|x>*; ie,

(4) a<x|x> = a*<x|x>; ie,

(5) a = a*; ie, a is real (zero being excluded as trivial).

Thus, in the example that I gave: bn.an is real (because each factor is real).